example: let f(x,y,z)=(yln(z), xln(z), xy/z) be a vector field. determine if the vector field is conservative.

Explanation:

Step1: Recall the curl - test for conservative fields

A vector field $\vec{F}(x,y,z)=\langle P(x,y,z),Q(x,y,z),R(x,y,z)\rangle$ is conservative if and only if $\nabla\times\vec{F}=\vec{0}$, where $\nabla\times\vec{F}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\P&Q&R\end{vmatrix}$. Here, $P = y\ln(z)$, $Q=x\ln(z)$, and $R=\frac{xy}{z}$.

Step2: Calculate the $i$ - component of the curl

The $i$ - component of $\nabla\times\vec{F}$ is $\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)$. $\frac{\partial R}{\partial y}=\frac{\partial}{\partial y}\left(\frac{xy}{z}\right)=\frac{x}{z}$, and $\frac{\partial Q}{\partial z}=\frac{\partial}{\partial z}(x\ln(z))=\frac{x}{z}$. So, $\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}=\frac{x}{z}-\frac{x}{z} = 0$.

Step3: Calculate the $j$ - component of the curl

The $j$ - component of $\nabla\times\vec{F}$ is $\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)$. $\frac{\partial P}{\partial z}=\frac{\partial}{\partial z}(y\ln(z))=\frac{y}{z}$, and $\frac{\partial R}{\partial x}=\frac{\partial}{\partial x}\left(\frac{xy}{z}\right)=\frac{y}{z}$. So, $\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}=\frac{y}{z}-\frac{y}{z}=0$.

Step4: Calculate the $k$ - component of the curl

The $k$ - component of $\nabla\times\vec{F}$ is $\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)$. $\frac{\partial Q}{\partial x}=\frac{\partial}{\partial x}(x\ln(z))=\ln(z)$, and $\frac{\partial P}{\partial y}=\frac{\partial}{\partial y}(y\ln(z))=\ln(z)$. So, $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=\ln(z)-\ln(z) = 0$. Since $\nabla\times\vec{F}=\langle0,0,0\rangle$, the vector field $\vec{F}$ is conservative.

Answer:

The vector field is conservative.